MCQMediumJEE Main 2025 · 3 April, Shift 1Continuity

Mathematics Question from JEE Main 2025 · 3 April, Shift 1

Let f(x)={(1+ax)1/x,x<01+b,x=0(x+4)1/22(x+c)1/32,x>0f(x) = \begin{cases} (1+ax)^{1/x} &, x < 0\\ 1+b &, x = 0\\ \frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} &, x > 0 \end{cases} be continuous at x=0x = 0. Then eabce^a bc is equal to

  • A

    6464

  • B

    7272

  • C

    4848

  • D

    3636

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